Suppose an equation that describes the path of a diver when diving off a platform is d = -5t^2 + 10t + 20. Where d is the distance above the water in (feet) and t is the time from the beginning of the dive(seconds). After how many seconds is the diver 25 feet above the water ? After how many seconds does the diver enter the water?

Answer 1

Solve for t in d(t)=25 and d(t)=0, where d(t) is given as:
d(t)= -5t^2 + 10t + 20

we find that d(1)=25, and d(3.23)=0.

Note: not sure if the diver was on earth. The acceleration due to gravity is only 10 instead of 32.2 f/s2 unless the distance d is in metres.